What is the integral of x^9\sqrt[3]{x^{10}+4} ?

The integral of x^9\sqrt[3]{x^{10}+4} is 3/40 (x^{10}+4)^{4/3}+C

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integral of x^9\sqrt[3]{x^{10}+4}

\frac{3}{40} (x^{10} + 4)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{10} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x^{10} + 4

= \frac{1}{10} \cdot \frac{3}{4} (x^{10} + 4)^{\frac{4}{3}}

Simplify

\frac{1}{10} \cdot \frac{3}{4} (x^{10} + 4)^{\frac{4}{3}} : \frac{3}{40} (x^{10} + 4)^{\frac{4}{3}}

= \frac{3}{40} (x^{10} + 4)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{40} (x^{10} + 4)^{\frac{4}{3}} + C

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What is the integral of x^9\sqrt[3]{x^{10}+4} ?
The integral of x^9\sqrt[3]{x^{10}+4} is 3/40 (x^{10}+4)^{4/3}+C